How High Does a Basketball Bounce When Thrown at an Angle?

[ryno2107]

An athlete throws a basketball upward from the ground, giving it speed 9.7 m/s at an angle of 64.0° above the horizontal.

(a) What is the acceleration of the basketball at the highest point in its trajectory?
m/s2

(b) On its way down, the basketball hits the rim of the basket, 3.05 m above the floor. It bounces straight up with one-half the speed with which it hit the rim. What height above the floor does the basketball reach on this bounce?

(a) is -9.8m/s2.

For (b) I found the time when the ball hits the rim 1.3s. Then I found the final velocity -3.04m/s and divided by 2 to get -1.52m/s which is the initial velocity of the ball when it bounces straight up vertically from the rim.

I am lost at where to go now and need help with solving (b).

 

[rl.bhat]

Your time is correct.
Check the calculation of the vertical component of the velocity at that time.

 

[tomcue]

For (b), you can use the equation for displacement in the vertical direction: y = y0 + v0t + 1/2at^2, where y is the final height, y0 is the initial height (3.05 m in this case), v0 is the initial velocity (1.52 m/s upward in this case), t is the time (which you already found to be 1.3 s), and a is the acceleration (-9.8 m/s2 in this case).

Substituting these values into the equation, we get:

y = 3.05 + (1.52)(1.3) + 1/2(-9.8)(1.3)^2

y = 3.05 + 1.976 - 8.058

y = -3.032 m

Therefore, the basketball reaches a height of 3.032 meters above the floor on its bounce.